Answer
$1$
Work Step by Step
Use hyperbolic functions formula as follows: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$ and $ \sinh x= \dfrac{e^x-e^{-x}}{2}$
Need to solve:$\cosh^2 x-\sinh^2x$
This implies: $\cosh^2 x-\sinh^2x=(\cosh x -\sinh x) (\cosh x +\sinh x)=[(\dfrac{e^x+e^{-x}}{2})-(\dfrac{e^x-e^{-x}}{2})][(\dfrac{e^x+e^{-x}}{2})-(\dfrac{e^x-e^{-x}}{2})]=(e^x) \cdot (e^{-x})=e^{x-x}=e^{0}$
Hence, $\cosh^2 x-\sinh^2x=1$
